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Miller Effect Cascode BJT Amplifier

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Miller Effect Cascode BJT Amplifier ESE319 Introduction to Microelectronics Miller Effect Cascode BJT Amplifier 2008 Kenneth R. Laker,update 18Oct10 KRL 1 ESE319 Introduction to Microelectronics Prototype Common Emitter Circuit Ignore “low frequency” High frequency model capacitors 2008 Kenneth R. Laker,update 18Oct10 KRL 2 ESE319 Introduction to Microelectronics Multisim Simulation Mid-band gain ∣Av∣=g m RC=40mS ∗5.1k =20446.2 dB Half-gain point 2008 Kenneth R. Laker,update 18Oct10 KRL 3 ESE319 Introduction to Microelectronics Introducing the Miller Effect The feedback connection of C between base and collector causes it to appear to the amplifier like a large capacitor 1 − K C has been inserted between the base and emitter terminals. This phenomenon is called the “Miller effect” and the capacitive multiplier “1 – K ” acting on C equals the common emitter amplifier mid-band gain, i.e. K = − g m R C . NOTE: Common base and common collector amplifiers do not suffer from the Miller effect, since in these amplifiers, one side of C is connected directly to ground. 2008 Kenneth R. Laker,update 18Oct10 KRL 4 ESE319 Introduction to Microelectronics High Frequency CC and CB Models B C E ground B v-pi C B v-pi C E E C Common Collector Common Base C is in parallel with R . C is in parallel with R . B C 2008 Kenneth R. Laker,update 18Oct10 KRL 5 ESE319 Introduction to Microelectronics Miller's Theorem I Z −I I 1=I I 2=−I + + + + V 1 K V 1 V 2 V 1 Z Z V 2=K V 1 <=> 1 2 - - - - V 1 V 1 Z V 1−V 2 V 1−K V 1 V 1 Z I = = = => 1= = = Z Z Z I 1 I 1−K 1−K 1 Z 1= 1 j 2 f C 1−K V 2− V 2 V 2−V 2 K V 2 −I = = = V 2 V 2 Z Z Z Z Z = = = ≈ Z => 2 I −I 1 if K >> 1 1 2 1− 1− K K approx. V 2 V 2 1 Z = = ≈ open 2 I I j 2 f C 2 − circuit 2008 Kenneth R. Laker,update 18Oct10 KRL 6 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis I Determine effect of C : C B V C V Using phasor notation: o I =−g V I I RC m C RC or g V V o= −g m V I C RC m where E I C = V −V o sC Note: The current through C depends only on V ! V o 2008 Kenneth R. Laker,update 18Oct10 KRL 7 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis II From slide 7: I C = V g m V RC −I C RC s C Collect terms for I and V : C 1s RC C I C = 1g m RC sC V 1g R sC s 1g R C I m C V m C V C = = 1s RC C 1s RC C Miller Capacitance C : C eq=1−K C =1g m RC eq 2008 Kenneth R. Laker,update 18Oct10 KRL 8 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis III C eq=1g m RC C For our example circuit: 1g m RC=10.040⋅5100=205 Ceq=205⋅2 pF≈410 pF 2008 Kenneth R. Laker,update 18Oct10 KRL 9 ESE319 Introduction to Microelectronics Apply Miller's Theorem to BJT CE Amplifier −I I 2=−I Z Z Z Z 2= 1= 1 1−K 1− K 1 For the BJT CE Amplifier: Z = and K=−gm RC j C 1 1 1 => Z 1= and Z 2= ≈ j 1g m RC C 1 j C j 1 C g m RC C eq=1g m RC C Miller's Theorem => important simplification to the HF BJT CE Model 2008 Kenneth R. Laker,update 18Oct10 KRL 10 ESE319 Introduction to Microelectronics Simplified HF Model R C s C V R r V r R R V s B o C L o g m V ' RL ' R =r ∥R ∥R ' I C L o C L C Rs ' RB∥r V V '' ' s= s V V C R V RB∥rRsig s L o g m V ' Rs=r∥ RB∥Rs Thevenin 2008 Kenneth R. Laker,update 18Oct10 KRL 11 ESE319 Introduction to Microelectronics Simplified HF Model ' V s Miller's Theorem ' I C Rs C in=C C eq ' ' V V s C C eq RL o ' .=C C 1g m RL g V C m (c) in 2008 Kenneth R. Laker,update 18Oct10 KRL 12 ESE319 Introduction to Microelectronics Simplified HF Model Note that Z = 1/jωC 2 µ is ignored! Why? ' V s V o dB V ∣ s∣ ' V o −gm RL Av f = ≈ ' V s 1 j 2 f C in Rs 1 f H = ' 2C in Rs 2008 Kenneth R. Laker,update 18Oct10 KRL 13 ESE319 Introduction to Microelectronics The Cascode Amplifier ● A two transistor amplifier used to obtain simultaneously: 1. Reasonably high input impedance. 2. Reasonable voltage gain. 3. Wide bandwidth. ● None of the conventional single transistor designs will satisfy all of the criteria above. ● The cascode amplifier will satisfy all of these criteria. ● A cascode is a combination of a common emitter stage cascaded with a common base stage. (Historical Note: the cascode amplifier was a cascade of grounded cathode and grounded grid vacuum tube stages – hence the name “cascode,” which has remained in modern terminology.) 2008 Kenneth R. Laker,update 18Oct10 KRL 14 ESE319 Introduction to Microelectronics The Cascode Circuit CB Stage CE Stage CB Stage i R C1 R 1 C ic2 C byp i B1 C1 v-out vo B1 vO R ib2 s ie1 ic1 i E1 E1 CE Stage i R iC2 ie2 b1 R C 2i V CC v R s in B2 C2 s R C R E B2 B E2 i E2 v v v s R R R R R eg1 cg2 low 3 R B= 2∥ 3 in1= = = E ie1 ic2 ac equivalent circuit Comments: 1. R1, R2, R3, and RC set the bias levels for both Q1 and Q2. 2. Determine RE for the desired voltage gain. 3. Cin and Cbyp are to act as “open circuits” at dc and act as “short circuits” at all operating frequencies of interest, i.e. f f min . 2008 Kenneth R. Laker,update 18Oct10 KRL 15 ESE319 Introduction to Microelectronics Cascode Mid-Band Small Signal Model RB=R2∥R3 2008 Kenneth R. Laker,update 18Oct10 KRL 16 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis ic1 1. Show reduction in Miller effect CB B1 vo 2. Evaluate small-signal voltage gain Stage C1 ib1 r v 1 g v 1 m 1 OBSERVATIONS E1 a. The emitter current of the CB stage is B2 ie1 R =low C2 ic2 in1 the collector current of the CE stage. (This R ib2 R s v g v C also holds for the dc bias current.) r 2 m 2 v 2 i i s E2 e1= c2 CE R ie2 R B E b. The base current of the CB stage is: Stage i i i = e1 = c2 b1 1 1 g m1=g m2=g m c. Hence, both stages have about same r r r e1= e2= e collector current i c1 ≈ i c2 and same g , r , r . m e π r1=r 2=r 2008 Kenneth R. Laker,update 18Oct10 KRL 17 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. The input resistance R to the CB stage is in1 the small-signal “r ” for the CE stage, i.e. e1 v g v ie1 ic2 1 m 1 i = = b1 1 1 The CE output voltage, the voltage drop from Q2 collector to ground, is: v 2 g m v2 r 1 r 1 v =v =−r i =− i =− i c2 e1 1 b1 1 c2 1 e1 Therefore, the CB Stage input resistance is: ve1 r1 Rin1= = =r e1 −ie1 1 v R r r A c2 in1 e 1 e vCE−Stage= ≈− =− => C eq=1 C 2C vs RE RE RE 2008 Kenneth R. Laker,update 18Oct10 KRL 18 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont. Now, find the CE collector current in terms of the input voltage v : i i s Recall c1≈ c2 v be11 g m v1 vs ib2≈ Rs∥RBr21 RE v v v s s vbe2 2 g m v2 ic2=ib2≈ ≈ Rs∥RBr 21 RE 1 RE for bias insensitivity: 1RE ≫Rs∥RBr v s vo −RC Av= = OBSERVATIONS: v s RE 1. Voltage gain A is about the same as a stand-along CE Amplifier. v 2. HF cutoff is much higher then a CE Amplifier due to the reduced C . eq 2008 Kenneth R. Laker,update 18Oct10 KRL 19 ESE319 Introduction to Microelectronics Approximate Cascode HF Voltage Gain C µ2 v 1 g m v1 RC C − µ1 V o f RE Av f = ≈ ' V s f 1 j 2 f C in Rs v 2 g m v2 where r C =C C =C 1 e C C 2C in eq R ' E Rs=Rs∥RB≈Rs 2008 Kenneth R. Laker,update 18Oct10 KRL 20 ESE319 Introduction to Microelectronics Cascode Biasing 1. Choose IE1 – make it relatively large to reduce R in 1 = r e1 = V T / I E1 to push out HF break frequencies. o.c. IC1 I 1 v C B O 2. Choose RC for suitable voltage swing R =low V and R for desired gain. I in1 C1 E E1 I C in C2 3.
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